 So on a previous video, that if a function has an absolute maximum or an absolute minimum, that those maximum or minimum values would be unique. But how do we know if they in exist or not? Because we saw some examples that they didn't exist. The extreme value theorem does come into play here, and it does give us conditions that will guarantee the existence of absolute maximum or absolute minimum values. That is, the extreme value theorem guarantees the existence absolute extrema. So if a function f is continuous, remember continuous means that the function is equal to its limit on all places in the domain. So if a function f is continuous on the closed interval a to b, so the closed interval means the points a and b are included inside of that interval, then this function will have an absolute maximum and an absolute minimum on that interval. And the basis of the idea comes from this. Let's draw like a continuous function and thinking of the domain here, let's pick some value. We have a and we have b. And so let's look at these values. So we have some f of a, we have some f of b. And so if we look at those, we look at just this little strip here on the graph. So looking just between these two lines, is there a place where it gets the biggest or someplace where it gets the smallest? And the answer is yes. The absolute largest value appears to be right here. We're actually touching the line. And the minimum value is likewise going to be happening here as well. Maybe like right here is the absolute minimum. Turns out there is more than one point that obtains the absolute minimum value there. And so if you have a closed interval on the continuous function, then the extreme value theorem guarantees that you will have an absolute extreme of both a maximum and a minimum. Now I should mention that the assumptions of the extreme value theorem are important. The reason we chose the assumptions the way we did is because if we relax any of the assumptions, then it could be a false statement then. So why is continuity important? Let's suppose we consider the function, let's just take that's standard tangent function y equals tangent of x and let's take the open interval negative pi halves to pi halves. I'll do continuity in just a second. So let's start with it. Why does it matter if it's a closed interval? Well, if you take an open interval, notice what happens on our graph, right? We get arbitrarily close to pi halves, but we never actually touch pi halves. And because that coincides with the vertical asymptote, the function gets bigger and bigger and bigger, it has no absolute maximum value. And because we can get arbitrarily close to negative pi halves, we get closer and closer to the asymptote, we're going to go off towards negative infinity. We see that there's no minimum value. There's no one number for which you can't get passed, right? You can get closer and closer and infinity and negative infinity without bound. So if you have an open interval, you can basically squeeze in some type of very large behavior like a vertical asymptote or whatever. It's necessary. We need to have a closed interval to guarantee that we have absolute extrema. But also, we need to be continuous. This is why I'm using this tangent example. Let's say we have a closed interval now. Let's go from 0 to pi. That's a closed interval. We include 0. That's part of the domain. We include pi as part of the domain. But because we have a discontinuity, particularly in this case, a vertical asymptote at x equals pi halves, there's no absolute maximum because we can get arbitrarily close to infinity. There's no absolute minimum because we can arbitrarily close to negative infinity. So the continuity assumption and the closed interval assumption are necessary in order to guarantee that we have absolute extrema on our function. So with the extreme value theorem in hand, we actually have a strategy for finding the absolute extrema for a function that's continuous on a closed interval. And so this is the stratagem that we're going to follow here. So if we're going to find an extreme value, we've saw with Fermat's theorem that the extreme values occur in one or two places. They occur at critical numbers or they occur at end points. Now, the end points are going to be pretty obvious. They're given to us a and b. But the critical numbers of the function between a and b won't exactly be obvious. So we have to search for them. So the first thing we have to do is we have to find all of the critical numbers of f. And that typically requires we're going to have to compute f prime of x. We have to calculate the first derivative because we have to identify where's the first derivative of zero or where is it does not exist. I'm not saying f prime of x equals does not exist because you can't equal d and e because that means it doesn't exist. But we have to figure out where is the derivative equal to zero horizontal tangent lines or where might the derivative be undefined if at all possible. So that finds us the critical numbers. We didn't evaluate the critical numbers. We figured out their y coordinates. We figured out the y coordinates of the end points. And then amongst this list, if we look at the y coordinates of the end points, the y coordinates of the critical numbers, the very biggest y coordinate would be the absolute maximum guaranteed by the extreme value theorem. And the very smallest value in that list would be the absolute minimum value guaranteed by the extreme value theorem. Let me show you an example of how this works. So let's consider the function f of x equals x to the eighth thirds power minus 16 x to the two thirds. And let's consider the interval negative one to eight. What are the absolute extreme of this function on that domain from negative one to eight? The extreme value theorem guarantees there is a solution here. We have a continuous function and we have a closed interval negative one to eight. So we have to identify what point should we be looking at. We first look at the derivative. So f prime of x by the usual power rule, we're going to get eight thirds x lower the power by one. So we're going to subtract one from eight thirds or subtract three thirds from eight thirds. This is going to give us five thirds minus 16 times two thirds lower the power by one. We're going to negative one third right there. We have to identify when this is thing equal to zero or when is it undefined? Okay, the best way to do this is to write this as a fraction. We're going to want to factor this thing is essentially what we need to do right now. So noticing some things we can do. 18, excuse me, 16 is factors as eight times two. So we can take out an eight. We can also take out a one third. So I'm going to take away the coefficient of the fraction eight thirds there. That leaves behind x to the five thirds minus we're going to have a four times x to negative one third. When you're trying to factor out powers of x, especially when there are fractions here, the best principle here that I can mention is factor away the smallest power. So in this case, we're going to factor away a negative one third power. The sign does matter here. We're factoring a negative one third power of x. So we get eight thirds times x to the negative one third and we're going to multiply that by what's left behind. What when you factor? Factoring is just division, but instead of throwing away your divisor at the end, you keep the factor. So we remember there's an x to the negative one third, but factoring just means divided. So we have to take x to the five thirds divided by x to the negative one third, right? And then we have minus four x to the negative one third divided by x to the negative one third. Division and factor in the same thing. It's just you remember what you divided by when you factor. x to the negative one third is going to cancel each other here. In terms of in terms of divided by x to the negative one third, we're really just attracting the exponent, which really since it's a negative, we're adding the exponent. We end up with this eight x to the negative one third over three. Then we're going to get x to the six thirds power minus four, which six thirds, of course, is the is just going to be the x squared there. So we end up with eight times x squared minus four. Notice how I didn't mention x to the negative one third that time. Well, it's because you have a negative x one that really could go in the denominator. So we get three times x to the one third and then x squared minus four. If I factor that as a difference of squares, I'm going to end up with eight times x minus two times x plus two all over three times x to the one third. So you'll notice that I took my derivative, I factored it and wrote it as a fraction. Why does that matter? Because why factor is important because if we want the derivative to be zero, that happens when the numerator goes to zero. The only way that a fraction equals zero is the numerator goes to zero. And so the numerator going to zero is going to determine for us where are the horizontal tangent lines of the function. On the other hand, what makes the denominator go to zero will make the derivative undefined. So f prime of x is d and e. But with a fraction, if the derivative goes to zero, excuse me, the denominator goes to zero, this will suggest that we have some type of vertical tangent, which of course the derivative is undefined at that location. So we have to figure out what makes the numerator go to zero, what makes the denominator go to zero. The numerator, since it's factored eight times x minus two and x plus two, we get the critical numbers of two and negative two. And in the denominator, if the denominator goes to zero, three x to the one third, that would suggest that x equals zero. So the derivative is undefined at x equals zero, and the derivative is zero at two and negative two. But all three of these numbers are our critical numbers right here. So these are our critical numbers. But one thing that's important to remember about critical numbers is the critical numbers need to be inside of the domain here, negative one to eight, which you'll notice that two and zero are between negative one and eight, but negative two is not. So we actually have to throw that number out. Because it's outside of our domain, it's infeasible. So it's whatever it does, doesn't matter, it has nothing to do with our problem, we have to throw it out. The function does have critical numbers outside of negative one and eight, but we only care about the extremum inside of the domain, negative one to eight. The next thing we're going to do is we're going to develop a t chart here, where on the left hand side, we're going to put our critical numbers and boundary points for x. So that'll include all those boundary points and the critical numbers we found. So we get negative one, which was the left boundary point, we get zero, which is a critical number, we get two, which is a critical number, and we get eight, which was a boundary point, we need to list those. Then we're going to look at f of x on the right hand side of our t chart. So we're going to evaluate the function, not the derivative, we're going to evaluate the function at these values. So we have to look at f of negative one, which remember our function was originally, but you can't see it on the screen now. Let me write it right here. So we're going to get x to the eighth thirds power minus 16 x to the two thirds, like so. And there are rational exponents, we could calculate these certainly, but you could use a calculator to help you, if you need to. So we have negative one to the eighth thirds power minus 16 times negative one, excuse me, to the negative one, two thirds power, taking the cube root of negative one is just negative one. So you're going to get negative one to the eighth power, which is actually a positive one. And then negative one squared is also going to be a positive one. So you get one minus 16, which is equal to negative 15. That's a value here to remember. And so honestly, I'm going to simplify this table and just write down the negative one 15, a negative 15, excuse me, like so. But like I said, we're trying to plug in, we're putting these into the function f of x, x to the eighth thirds minus 16 x to the two thirds, plug it in zero is pretty easy. You're going to get zeros everywhere. So that gives you a zero eights. Of course, a nice choice because eight's a perfect cube. So when you plug in eight f of eight here, you're going to end up with eight to the eight thirds minus 16 to the eight thirds, to two thirds, excuse me, when you take the cube root of eight, you're going to get two. So you get two to the eighth minus 16 times two to the square, like so. In the end, this will simplify because I mean two to the eighth, what's that going to be? That's 256. And then you're going to subtract from 256, 64. So you get 192. And then lastly, you have to do f of two, which again, this one does the two is not a perfect cube like the other numbers were. So this one, you pretty much need to have your calculator to get a good approximation of what's going on here. But you're going to plug this number in here. And that's going to give you approximately negative 19.05 if I round it to two decimal places. So this here is just kind of like scratch work. Don't worry too much about it. Because that's just how we calculate these things. This table is what matters. When you look at the table, the very biggest value on the right side of the table will give you the absolute maximum. So you see this right here, we're looking at the y-coordinate here. This is the absolute maximum value of the graph because 192 is the biggest value you get. No one else is bigger. So that's going to be the absolute maximum. And the maximum occurs at x equals eight. The maximum value itself is 192. Where is the minimum value? The absolute minimum happens here at x equals two. The absolute minimum value, the y-coordinate, is going to be negative 19.05 and it's obtained at x equals two. So this is a nice thing about the extreme value theorem. When you combine that with Fermat's theorem, because the extreme value theorem says on a closed interval, you're guaranteed to have absolute maximum, absolute minimum values. And Fermat's theorem tells you that they have to be at end points or critical numbers. You'll notice that the absolute maximum occurred at a boundary point, x equals eight, and the minimum happened at the critical number two, which is where the derivative had a horizontal tangent line. And so this demonstrates how one can solve the so-called extreme value problem. If you want to find the extreme values on a continuous function on a closed interval, you calculate the critical numbers, plug those into the function, not the derivative, plug the critical numbers and the end points into the function. And then the biggest value present will be the absolute maximum, the smallest value will be the absolute minimum. And that's all there is to it.